Block on an Inclined Plane

A block on a tilted ramp. Gravity pulls down, friction resists, and an applied force can push uphill. Tilt past the friction angle and gravity wins.

Visualization

Input

\theta

Ramp angle

\mathrm{deg}

m

Mass

\mathrm{kg}

\mu_s

Static friction coefficient

\mu_k

Kinetic friction coefficient

F

Applied force (up-ramp)

\mathrm{N}

g

Gravity

\mathrm{m/s^{2}}

Show the free-body diagram

Splitting gravity onto the slope

A block sitting on a tilted ramp is pulled straight down by gravity, but the ramp is in the way. Only part of gravity actually tries to slide the block down the slope. The other part just presses the block into the ramp surface.

Splitting gravity into those two directions is the whole trick. With ramp angle $\theta$ :

W_\parallel = mg \sin\theta \qquad W_\perp = mg \cos\theta

The first piece, $W_\parallel$ , is what actually pulls the block down-slope. The second piece, $W_\perp$ , gets canceled by the ramp pushing back on the block — that's the normal force, $N$ . Since the block isn't sinking into the ramp, the two have to match exactly:

N = W_\perp = mg \cos\theta

Source: OpenStax Physics §5.4 — Inclined Planes.

Friction has two modes

Friction is the interesting part of this whole problem. It behaves completely differently depending on whether the block is sitting still or sliding.

Static — block isn't moving. Friction is reactive: it pushes back with whatever force is needed to prevent motion, up to a ceiling set by the materials in contact:

f_s \le \mu_s N

Push the block harder than that ceiling and friction can't keep up. The block starts to slide.

Kinetic — block is sliding. Friction now has a fixed value, smaller than the static ceiling, that always opposes the motion:

f_k = \mu_k N

The two coefficients $\mu_s$ and $\mu_k$ are properties of the surfaces in contact — wood on wood, rubber on concrete, shoes on ice. The OpenStax friction page has a table of typical values. OpenStax flags upfront that this whole picture "is an approximate empirical model only" — it works well for everyday dry surfaces and breaks down for lubricated contacts or very high speeds.

Putting it together

The non-friction force pulling the block down-slope is whatever gravity gives ( $W_\parallel$ ) minus whatever you push back with ( $F$ ):

F_\text{drive} = W_\parallel - F

If that's still under the static ceiling — $|F_\text{drive}| \le \mu_s N$ — then static friction matches it and the block stays put. Otherwise the block breaks free and slides. Kinetic friction takes a fixed bite, and what's left accelerates the block by Newton's second law:

a = \frac{F_\text{drive} - \mu_k N}{m}

Source: Wikipedia — Newton's laws of motion.

Show your work

Weight: $W \;=\; m g \;=\; \textcolor{#dc4124}{\left(5\right)} \textcolor{#dc4124}{\left(9.81\right)} \;=\; 49.05\;\mathrm{N}$
Weight component along ramp (down-slope): $W_{\parallel} \;=\; m g \sin\theta \;=\; \textcolor{#dc4124}{\left(5\right)} \textcolor{#dc4124}{\left(9.81\right)} \sin\theta \;=\; 24.52\;\mathrm{N}$
Weight component normal to ramp: $W_{\perp} \;=\; m g \cos\theta \;=\; \textcolor{#dc4124}{\left(5\right)} \textcolor{#dc4124}{\left(9.81\right)} \cos\theta \;=\; 42.48\;\mathrm{N}$
Normal force: $N \;=\; W_{\perp} \;=\; 42.48\;\mathrm{N}$
Maximum static friction: $F_{s,\max} \;=\; \mu_s N \;=\; \mu_s \textcolor{#dc4124}{\left(42.48\right)} \;=\; 16.99\;\mathrm{N}$
Net non-friction force along ramp (down-slope positive): $F_{\text{drive}} \;=\; W_{\parallel} - F \;=\; W_{\parallel} - \textcolor{#dc4124}{\left(0\right)} \;=\; 24.52\;\mathrm{N}$
Friction force (up-slope when sliding down, down-slope when sliding up): $F_f \;=\; \begin{cases} F_{\text{drive}} & |F_{\text{drive}}| \le F_{s,\max} \\ \operatorname{sign}(F_{\text{drive}})\,\mu_k N & \text{otherwise} \end{cases} \;=\; 12.74\;\mathrm{N}$
Net force on block along ramp (down-slope positive): $F_{\text{net}} \;=\; F_{\text{drive}} - F_f \;=\; 11.78\;\mathrm{N}$
Acceleration along ramp (down-slope positive): $a \;=\; \frac{F_{\text{net}}}{m} \;=\; \frac{F_{\text{net}}}{\textcolor{#dc4124}{\left(5\right)}} \;=\; 2.356\;\mathrm{m/s^{2}}$

Worked Example

OpenStax University Physics Vol. 1, Example 6.14: a snowboarder is sliding down a 13° slope with kinetic friction coefficient μ_k = 0.20. The source reports the resulting acceleration as 0.29 m/s² down the slope. Mass cancels in this problem (a = g(sin θ − μ_k cos θ)); here we set m = 70 kg as a representative value and μ_s = μ_k since the source does not specify μ_s but states the snowboarder is sliding.

Quantity	Expected	Computed	Δ
a	0.29	0.2948	0.004755
is_static	0	0	0

Source: OpenStax, University Physics Vol. 1, §6.2 Example 6.14 (Snowboarder)

Sources

[1] OpenStax. Physics — §5.4 Inclined Planes, eq. w_‖ = mg sin θ; w_⊥ = mg cos θ; N = w_⊥ (openstax.org)
[2] OpenStax. University Physics Volume 1 — §6.2 Friction, eq. f_s ≤ μ_s N; f_k = μ_k N (Coulomb friction, approximate empirical model) (openstax.org)
[3] Wikipedia — Newton's laws of motion (Second Law: F = m a) (en.wikipedia.org)
[4] Wikipedia — Friction (dry friction, Coulomb model) (en.wikipedia.org)

Assumptions

Coulomb friction model — "an approximate empirical model only"; not accurate for lubricated surfaces or high-speed sliding (OpenStax UP Vol. 1 §6.2).
Friction acts parallel to the contact surfaces and opposes motion or impending motion (OpenStax UP Vol. 1 §6.2).
Weight decomposes onto the slope as w_‖ = mg sin θ and w_⊥ = mg cos θ; the normal force balances w_⊥ (OpenStax Physics §5.4).
Applied force F is taken parallel to the ramp surface (problem framing here; not a source assumption).